Ex-18.1, Basic Geometrical Tools, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics PDF Download

1.) Construct the following angles using set- squares:

Answer:

(i) 45^o

Place 45^o set- square.

Draw two rays AB and AC along the edges from the vertex from the vertex of 45^o angle of the set- square.

The angle so formed is a 45^o angle.

∠BAC=45^o

(ii) 90^o

Place = 90^o set –square as shown in the figure.

Draw two rays BC and BA along the edges from the vertex of 90^o angle.

The angle so formed is 90^o angle.

∠ABC=90^o

(iii) 60^o

Place 30^o set –square as shown in the figure.

Draw the rays BA and BC along the edges from the vertex of 60^o

The angle so formed is 60^o

∠ABC=60^o

(iv) 105^o

Place 30^o set –square and make an angle 60^o by drawing the rays BA and BC as shown in figure.

Now place the vertex of 45^o of the set –square on the ray BA as shown in figure and draw the ray BD.

The angle so formed is 105^o

Therefore, ∠DBC=105^o

(v ) 75^o

Place 45^o set –square and make an angle of 45^o by drawing the rays BD and BC as shown in the figure.

Now place the vertex of 30^o of the set- square on the ray BD as shown in the figure and draw the ray BA.

The angle so formed is 75^o.

Therefore, ∠ABC=75^o

(Line BD is hidden)

(vi) 150^o

Place the vertex of 45^o of the set – square and make angle of 90^o by drawing the rays BD and BC as shown in the figure

Now, place the vertex of 30^o of the set –square on the ray BS as shown in the figure and draw the ray BA

The angle so formed is 150^o.

Therefore, ∠ABC=150^o

2.) Given a line BC and a point A on it, construct a ray AD using set – squares so that ∠DAC is

(i) 30^o

(ii) 150^o

Answer:

(i) Draw a line BC and take a point A on it. Place 30^o set –square on the line BC such that its vertex of 30^o angle lies on point A and one edge coincides with the ray AB as shown in figure

Draw the ray AD.

Thus ∠DAC is the required angle of 30^o

(ii) Draw a line BC and take a point A on it. Place 30^o set –square on the line BC such that its vertex of 30^o angle lies on point A and one edge coincides with the ray AB as shown in the figure.

Draw the ray AD.

Therefore, ∠DAB=30^o

We know that angle on one side of the straight line will always add to 180^o

Therefore, ∠DAB+∠DAC=180^o

Therefore, ∠DAC=150^o